The length of the pool is L = 20 m and its depth is D = 4 m at the deep end. It also has a width W = 5 m. The water line rises as water flows into the pool. At any time t the water line has a length l, and the water fills the pool to a depth d. The water also has a width w = W = 5 m.
The volume of water in the pool is V = 0.5 * l * d * w (the volume of a right triangular prism) = 2.5 * l * d (since w = 5 m), and the flow rate of water into the pool is dV/dt = 2 m^3 / min. The problem is asking for dd/dt when d = 1 m.
Start by noting that, by similar triangles, l / d = L / D = 20 m / 5 m = 4, so that l = 4d. Now the volume of water in the pool can be written entirely in terms of d: V = 2.5 * l * d = 2.5 * 4d * d = 10d^2.
Now we are ready to differentiate the volume equation implicitly, with respect to t, to find dd/dt.
dV/dt = dV/dd * dd/dt = 2 * 10d * dd/dt = 20d * dd/dt = 20 dd/dt when d = 1 m => dd/dt = 1 / 20 * dV/dt
= 1 / 20 * 2 = 0.1 m / min when d = 1 m..
Optional:
One can check the sense of this answer by noting that the top of the water has a surface area of
A = l * w = 4d * 5 m = 20 m^2 when d = 1 m. So, the 2 m^3 of water flowing into the pool in one minute, would spread over an area of 20 m^2 and, thus, must lead to a vertical rise of 0.1 m in the depth of the water in that minute.
Of course, the bottom of the pool is slanted, so the water doesn't rise vertically over the entire minute, but if the volume of incoming water continued spreading out in the same way it is at the instant d = 1 m, then it would need to rise a vertical height of 0.1 m.
John R.
04/30/19